Question

If $g(x)=1+\sqrt{x}$ and $f(g(x))=3+2\sqrt{x}+x$, then $f(x)$ is

• A. $1+2x^2$
• B. $2+x^2$
• C. $1+x$
• D. $2+x$

Answer 1

Answer: (B)

$f(1+\sqrt{x})=3+2\sqrt{x}+x$
Put $1+\sqrt{x}=y\Rightarrow f(y)=2+y^2$.